Square root of i

By Brad Rodgers (P1930) on Friday, April 28, 2000 - 03:01 am :

What is the square root of i?
I know it's not i, as then i=-1 and 1=-1.

Brad

By Dan Goodman (Dfmg2) on Friday, April 28, 2000 - 04:33 am :

Try this approach: you know the answer is a complex number of the form a+ib where a and b are real numbers, so you want to solve (a+ib)² =i. So you need to solve a² -b² =0 and 2ab=1. This gives the correct solution very quickly of sqrt(i)=±(1+i)/sqrt(2). This method (which I discussed in the thread "Imaginary Numbers" ) usually works for questions like this.

Another method is to use De Moivre's formula

(\cos (θ) + i \sin (θ {))}^{n} = \cos (n θ) + i \sin (n θ)

. Put

n = 2

and

θ = 45

and you get

(\cos 45 + i \sin 45)^{2} = \cos 90 + i \sin 90 = i

, so

\sqrt{i} = \cos 45 + i \sin 45

. You also know that

\cos^{2} x + \sin^{2} x = 1

, and

\cos 45 = \sin 45

(draw a picture), so

\cos 45 = \sin 45 = 1 / \sqrt{2}